Posted by Melody on Thursday, October 15, 2009 at 2:51am.
I am not too sure what you mean by "if we start the analysis one minute into the ride"
Where do you want the chair to be at 1 minute. The way it is, the chair would be 26.2 m high, (which also confirms my answer of 30 m at 1.09 minutes)
Do you want the chair to be at a minimum height of 2 m when the time is 1 minute?
If that is the case, we will have to perform a phase shift on our equation.
So we want h = 2 at t=1, rather than t=0
this will do it:
h = -35cos(2pi/5)(X-1) + 37
check: remember it takes 5 minutes for one rotation.
t = 1, h = 2 , at the bottom
t = 2.25 , h = 37, the height of the axle, check!
t = 3.5, h = 72 , the height of axle + radius of wheel, check!
t = 4.75, h = 37 , coming down, height of axle, check!
t = 6 , h = 2, back at the bottom, check!
In the original equation, using t values of 0, 1.25, 2.5, 3.75 and 5 will yield the same heights.
(That is how I checked if your equation was valid)
BTW, I gave you a time of 1.09 minutes for a height of 30 m.
That would be on the way up, of course you would reach another height of 30 m on your way down, which would be at (5-1.09) or 3.91 min.
repeatedly adding 5 minutes to these answers would produce all the times that the chair is 30 m high.
t =
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